Maximal systole of hyperbolic surface with largest $S^3$ extendable abelian symmetry
Geometric Topology
2023-09-06 v2 Differential Geometry
Abstract
We give the formula for the maximal systole of the surface admits the largest -extendable abelian group symmetry. The result we get is . Here \begin{eqnarray*} K &=& \sqrt[3]{\frac{1}{216}L^3 +\frac{1}{8} L^2 + \frac{5}{8} L - \frac{1}{8} + \sqrt{\frac{1}{108}L(L^2+18L+27)} } & & + \sqrt[3]{\frac{1}{216}L^3 +\frac{1}{8} L^2 + \frac{5}{8} L - \frac{1}{8} - \sqrt{\frac{1}{108}L(L^2+18L+27)} } & & + \frac{L+3}{6}. \end{eqnarray*} and .
Cite
@article{arxiv.1911.10474,
title = {Maximal systole of hyperbolic surface with largest $S^3$ extendable abelian symmetry},
author = {Yue Gao and Jiajun Wang},
journal= {arXiv preprint arXiv:1911.10474},
year = {2023}
}
Comments
38 pages, 46 figures